Termination w.r.t. Q proof of TRS/AProVErta3.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

Q restricted rewrite system:
The TRS R consists of the following rules:

ack₂(0, y) -> s₁(y)
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))
f₂(s₁(x), y) -> f₂(x, s₁(x))
f₂(x, s₁(y)) -> f₂(y, x)
f₂(x, y) -> ack₂(x, y)
ack₂(s₁(x), y) -> f₂(x, x)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

ack₂(0, y) -> s₁(y)
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))
f₂(s₁(x), y) -> f₂(x, s₁(x))
f₂(x, s₁(y)) -> f₂(y, x)
f₂(x, y) -> ack₂(x, y)
ack₂(s₁(x), y) -> f₂(x, x)

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

ACK₂(s₁(x), s₁(y)) -> ACK₂(s₁(x), y)
ACK₂(s₁(x), s₁(y)) -> ACK₂(x, ack₂(s₁(x), y))
F₂(s₁(x), y) -> F₂(x, s₁(x))
F₂(x, s₁(y)) -> F₂(y, x)
ACK₂(s₁(x), y) -> F₂(x, x)
F₂(x, y) -> ACK₂(x, y)
ACK₂(s₁(x), 0) -> ACK₂(x, s₁(0))

The TRS R consists of the following rules:

ack₂(0, y) -> s₁(y)
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))
f₂(s₁(x), y) -> f₂(x, s₁(x))
f₂(x, s₁(y)) -> f₂(y, x)
f₂(x, y) -> ack₂(x, y)
ack₂(s₁(x), y) -> f₂(x, x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

ACK₂(s₁(x), s₁(y)) -> ACK₂(s₁(x), y)
ACK₂(s₁(x), s₁(y)) -> ACK₂(x, ack₂(s₁(x), y))
F₂(s₁(x), y) -> F₂(x, s₁(x))
F₂(x, s₁(y)) -> F₂(y, x)
ACK₂(s₁(x), y) -> F₂(x, x)
F₂(x, y) -> ACK₂(x, y)
ACK₂(s₁(x), 0) -> ACK₂(x, s₁(0))

The TRS R consists of the following rules:

ack₂(0, y) -> s₁(y)
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))
f₂(s₁(x), y) -> f₂(x, s₁(x))
f₂(x, s₁(y)) -> f₂(y, x)
f₂(x, y) -> ack₂(x, y)
ack₂(s₁(x), y) -> f₂(x, x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

ACK₂(s₁(x), s₁(y)) -> ACK₂(x, ack₂(s₁(x), y))
F₂(s₁(x), y) -> F₂(x, s₁(x))
F₂(x, s₁(y)) -> F₂(y, x)
ACK₂(s₁(x), y) -> F₂(x, x)
ACK₂(s₁(x), 0) -> ACK₂(x, s₁(0))

The remaining pairs can at least be oriented weakly.

ACK₂(s₁(x), s₁(y)) -> ACK₂(s₁(x), y)
F₂(x, y) -> ACK₂(x, y)

Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(ACK₂(x₁, x₂)) = 2·x₁
POL(F₂(x₁, x₂)) = 2·x₁ + x₂
POL(ack₂(x₁, x₂)) = 0
POL(f₂(x₁, x₂)) = 0
POL(s₁(x₁)) = 2 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

ACK₂(s₁(x), s₁(y)) -> ACK₂(s₁(x), y)
F₂(x, y) -> ACK₂(x, y)

The TRS R consists of the following rules:

ack₂(0, y) -> s₁(y)
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))
f₂(s₁(x), y) -> f₂(x, s₁(x))
f₂(x, s₁(y)) -> f₂(y, x)
f₂(x, y) -> ack₂(x, y)
ack₂(s₁(x), y) -> f₂(x, x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

ACK₂(s₁(x), s₁(y)) -> ACK₂(s₁(x), y)

The TRS R consists of the following rules:

ack₂(0, y) -> s₁(y)
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))
f₂(s₁(x), y) -> f₂(x, s₁(x))
f₂(x, s₁(y)) -> f₂(y, x)
f₂(x, y) -> ack₂(x, y)
ack₂(s₁(x), y) -> f₂(x, x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

ACK₂(s₁(x), s₁(y)) -> ACK₂(s₁(x), y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ACK₂(x₁, x₂)) = 2·x₂
POL(s₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

ack₂(0, y) -> s₁(y)
ack₂(s₁(x), 0) -> ack₂(x, s₁(0))
ack₂(s₁(x), s₁(y)) -> ack₂(x, ack₂(s₁(x), y))
f₂(s₁(x), y) -> f₂(x, s₁(x))
f₂(x, s₁(y)) -> f₂(y, x)
f₂(x, y) -> ack₂(x, y)
ack₂(s₁(x), y) -> f₂(x, x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.